The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint : Sx – 1 = S49 – Sx ]
Row houses are numbers from 1,2,3,4,5…….49.
First term, a = 1
Common difference, d=1
Let us say the number of xth houses can be represented as;
Sum of nth term of AP = n/2[2a+(n-1)d]
Sum of number of houses beyond x house = Sx-1
= (x-1)/2[2.1+(x-1-1)1]
= (x-1)/2 [2+x-2]
= x(x-1)/2 ……………………(i)
By the given condition, we can write,
S49 – Sx
= \(\frac{49}{2}[2.1+(49-1)1]-\frac{x}{2}[2.1 + (x-1)1]\)
= 25(49) – x(x + 1)/2 ……………….(ii)
As per the given condition, eq.(i) and eq(ii) are equal to each other;
Therefore,
x(x-1)/2 = 25(49) – x(x-1)/2
x = ±35
As we know, the number of houses cannot be a negative number. Hence, the value of x is 35.